Actuarial Outpost SOA 289 #124
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#1
04-19-2010, 09:27 AM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FA, GH Core College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,176
SOA 289 #124

I understand the SOA's solution, but I am not sure as why mine doesn't work.

$Pr(X \ge 8000) = 0.5$, we agree there

$Pr(n \ge 2) = 1 - P(n = 0) - P(n=1) = 1-0.6^8-0.6^7(0.4)(8) = 0.892$

$Pr(X \ge 8000)^2 = 0.25$

$Pr(n \ge 2 \cap X \ge 8000) = Pr(n \ge 2)Pr(X \ge 8000) = 0.25*0.892 = 0.223$
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#2
04-19-2010, 09:57 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,464

You include the case that 3 scientists are eaten (as you should).

If three are eaten, what is the probility that exactly 2 of them exceed 8000 calories? You are using .25, and you shouldn't.

Similarly for 4,5,... scientists eaten.
#3
04-19-2010, 10:55 AM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FA, GH Core College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,176

Quote:
 Originally Posted by Gandalf You include the case that 3 scientists are eaten (as you should). If three are eaten, what is the probility that exactly 2 of them exceed 8000 calories? You are using .25, and you shouldn't. Similarly for 4,5,... scientists eaten.
The probability that one of them is eaten is the same as them not being eaten. Therefore, the probability that two of three are eaten is the same as none of them being eaten, or one, or three... $0.5^3$.

Thanks for the response Gandalf, but what am I missing?
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P, FM, MLC, MFE, C, FAPmods, APC, VEE, IA, HFmod, FHEmod, FA, GH Core, PRFmod, GH Adv, GH Spec, DMAC, FAC

How to explain actuarial exams to someone else...

Good Einstein quote - "One had to cram all this stuff into one's mind for the examinations, whether one liked it or not. This coercion had such a deterring effect on me that, after I had passed the final examination, I found the consideration of any scientific problems distasteful to me for an entire year."
#4
04-19-2010, 11:10 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,464

Quote:
 Originally Posted by colby2152 The probability that one of them is eaten is the same as them not being eaten. Therefore, the probability that two of three are eaten is the same as none of them being eaten, or one, or three... $0.5^3$. Thanks for the response Gandalf, but what am I missing?
That .5^3 for each case is wrong. By that logic:
p(0)+p(1)+p(2)+p(3) = 4/8. Where's the rest of the probability? exactly 2.5 exceed 8000? more than 3 of 3 exceed 8000?

The specific answer for the 3 eaten case is that p(2) = (3 choose 2)(.5^2)(.5) = .375.

The general problem with your approach is that you effectively assumed p(2) = .25 independent of how many were eaten. I don't think that approach - summing p(exactly n eaten)*p(exactly 2 over 8000 | exactly n eaten) - is feasible with a calculator, but it should work in Excel.
#5
04-19-2010, 11:20 AM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FA, GH Core College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,176

Hmm, I did disregard the combinations since the probability was 50%. Was that my problem?
__________________
P, FM, MLC, MFE, C, FAPmods, APC, VEE, IA, HFmod, FHEmod, FA, GH Core, PRFmod, GH Adv, GH Spec, DMAC, FAC

How to explain actuarial exams to someone else...

Good Einstein quote - "One had to cram all this stuff into one's mind for the examinations, whether one liked it or not. This coercion had such a deterring effect on me that, after I had passed the final examination, I found the consideration of any scientific problems distasteful to me for an entire year."
#6
04-19-2010, 11:31 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,464

Quote:
 Originally Posted by colby2152 $Pr(n \ge 2 \cap X \ge 8000) = Pr(n \ge 2)Pr(X \ge 8000) = 0.25*0.892 = 0.223$
The correct formula would be
$\sum_2^\infty Pr(N = n \cap 2out of n \ge 8000)$
Since $Pr(2 out of n \ge 8000)$ depends on n, you cannot convert it to a single value or take it outside the summation.

And you can't do the summation on an SOA calculator.
#7
04-20-2010, 10:22 PM
 jhp8 Member Join Date: Jan 2007 Posts: 278

Quote:
 Originally Posted by colby2152 I understand the SOA's solution, but I am not sure as why mine doesn't work. $Pr(X \ge 8000) = 0.5$, we agree there $Pr(n \ge 2) = 1 - P(n = 0) - P(n=1) = 1-0.6^8-0.6^7(0.4)(8) = 0.892$ $Pr(X \ge 8000)^2 = 0.25$ $Pr(n \ge 2 \cap X \ge 8000) = Pr(n \ge 2)Pr(X \ge 8000) = 0.25*0.892 = 0.223$
Not to hi-jack the thread, but I don't really understand the SOA answer either.

"Since each time the probability of a heavy scientist is just half the probability of a success, the distribution is binomial with q = 0.6×0.5 = 0.3 and m = 8."

I understand the idea of modifying the binomial, but I am not sure why we can apply this to this problem.
#8
04-20-2010, 10:29 PM
 ZeroUrashima Member Join Date: Apr 2010 Posts: 30

You just change the definition of "success".

The original N was the number of scientists eaten.
You change N to N* which is the number of scientists eaten AND above 8000 Calories.

So out of the original q=0.6, 0.3 is Above or Equal 8000 Calories, 0.3 is Below 8000 Calories. The 0.3 that is below 8000 calories would also be a a "failure" in the new distribution.

N* ~ Bino (q*=0.3,m=8)

Then Find P(N*=2)
#9
04-20-2010, 10:38 PM
 jhp8 Member Join Date: Jan 2007 Posts: 278

Quote:
 Originally Posted by ZeroUrashima You just change the definition of "success". The original N was the number of scientists eaten. You change N to N* which is the number of scientists eaten AND above 8000 Calories. So out of the original q=0.6, 0.3 is Above or Equal 8000 Calories, 0.3 is Below 8000 Calories. The 0.3 that is below 8000 calories would also be a a "failure" in the new distribution. N* ~ Bino (q*=0.3,m=8) Then Find P(N*=2)
Thanks....that really helped clear it up. Your explanation made me realize the piece I was missing was that point if two scientists are eaten and have > 8,000 calories, the first statement "Calculate the probability that two or more scientist are eaten" is automatically true and doesn't have to be accounted for.
#10
01-07-2011, 03:25 PM
 SDY2010 Join Date: Nov 2010 Posts: 6

I understand that q is modified to be 0.3.

The part of the solution that solves for Pr(N*=2) seems wrong. The question states that you want the probability that 2 or more scientists are eaten, not just 2.

Why would you not solve:
1-Pr(N*=0)-Pr(N*=1)
=1-(0.7^8)-[(8)*(0.3)*(0.7^7)]

 Tags allosaur, dinoz arr wierd, exam c, probability, soa 289

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